Exercises on orthogonal matrices and Gram-Schmidt Problem 17.1: (4.4 #10.b Introduction to Linear Algebra: Strang) Orthonormal vectors are automatically linearly independent. Matrix Proof: Show that Qx = 0 implies x = 0. Since Q may be rectan gular, you can use Q T but not Q−1 . Solution: By deﬁnition, Q is a matrix whose columns are orthonormal, and so we know that Q T Q = I (where Q may be rectangular). Then: Qx = 0 =⇒ Q T Qx = Q T 0 =⇒ Ix = 0 =⇒ x = 0. Thus the nullspace of Q is the zero vector, and so the columns of Q are linearly independent. There are no non-zero linear combinations of the columns that equal the zero vector. Thus, orthonormal vectors are auto matically linearly independent. Problem 17.2: (4.4 #18) Given the vectors a, b and c listed below, use the Gram-Schmidt process to ﬁnd orthogonal vectors A, B, and C that span the same space. a = (1, −1, 0, 0), b = (0, 1, −1, 0), c = (0, 0, 1, −1). Show that {A, B, C} and {a, b, c} are bases for the space of vectors per pendicular to d = (1, 1, 1, 1). Solution: We apply Gram-Schmidt to a, b, c. First, we set A = a = (1, −1, 0, 0). Next we ﬁnd B : AT b 1 B = b − T A = (0, 1, −1, 0) + (1, −1, 0, 0) = 2 A A � � 1 1 , , −1, 0 . 2 2 And then we ﬁnd C : AT c BT c 2 C = c − T A − T B = (0, 0, 1, −1) + 3 A A B B � 1 1 , , −1, 0 2 2 � � = � 1 1 1 , , , −1 . 3 3 3 We know from the ﬁrst problem that the elements of the set {A, B, C} are linearly independent, and each vector is orthogonal to (1,1,1,1). The space of vectors perpendicular to d is three dimensional (since the row space of (1, 1, 1, 1) is one-dimensional, and the number of dimensions of the row space added to the number of dimensions of the nullspace add to 4). Therefore {A, B, C} forms a basis for the space of vectors perpendicular to d. Similarly, {a, b, c} is a basis for the space of vectors perpendicular to d because the vectors are linearly independent, orthogonal to (1,1,1,1), and because there are three of them.